Question 466372
let sheep be {{{x}}} and gees {{{y}}}

sheep has 4 legs: {{{4x}}}

gees have 2 legs: {{{2y}}}

there are {{{46}}} legs: {{{4x+2y=46}}}


but only 14 heads: {{{x+y=14}}}

{{{4x+2y=46}}}
{{{x+y=14}}}
--------------------solve this system

{{{x+y=14}}}....solve for {{{x}}}

{{{x=14-y}}}......substitute in {{{4x+2y=46}}}


{{{4(14-y)+2y=46}}}


{{{56-4y+2y=46}}}


{{{56-2y=46}}}


{{{56-46=2y}}}

{{{10=2y}}}

{{{5=y}}}

{{{x=14-y}}

{{{x=14-5}}

{{{x=9}}


so, there are {{{9}}} sheep and {{{5}}} gees